Given that the mass of the Earth is $M$ and its radius is $R$. $A$ body is dropped from a height equal to the radius of the Earth above the surface of the Earth. When it reaches the ground,its velocity will be:

The radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth,then the change in potential energy will be:

Assuming that the gravitational potential energy of an object at infinity is zero,the change in potential energy (final - initial) of an object of mass $m$,when moved to a height $h$ from the surface of the Earth (of radius $R$),is given by:

Two uniform solid spheres of equal radii $R$,but mass $M$ and $4M$ have a centre-to-centre separation $6R$,as shown in the figure. The two spheres are held fixed. $A$ projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the centre of the second sphere. Obtain an expression for the minimum speed $v$ of the projectile so that it reaches the surface of the second sphere.

$A$ rocket is fired vertically with a speed of $5 \; km/s$ from the Earth's surface. How far from the Earth's surface does the rocket go before returning? (Mass of the Earth $M_e = 6.0 \times 10^{24} \; kg$,mean radius of the Earth $R_e = 6.4 \times 10^{6} \; m$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

$A$ projectile is thrown straight upward from the Earth's surface with an initial speed $v = \alpha v_E$,where $\alpha$ is a constant and $v_E$ is the escape speed. The projectile travels up to a height $h = 800 \ km$ from the Earth's surface before it comes to rest. The value of the constant $\alpha$ is (Radius of the Earth $R = 6400 \ km$):